Question 9 - A-Level Maths

CAIE FP1 2015 June — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2015
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.8 This is a second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with two distinct real roots) and particular integral (using trial solution for trigonometric forcing function), followed by applying two initial conditions to find constants. While systematic, it involves multiple substantial steps and careful algebraic manipulation, making it moderately challenging but still within standard Further Maths scope.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

9 Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 10 x = 2 \sin t - 3 \cos t$$ given that, when $t = 0 , x = 3.3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.9$.

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Question 9:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(m-5)(m+2) = 0 \Rightarrow \text{CF}: x = Ae^{5t} + Be^{-2t}$	M1A1
PI: $x = p\sin t + q\cos t \Rightarrow \dot{x} = p\cos t - q\sin t \Rightarrow \ddot{x} = -p\sin t - q\cos t$	M1
$\Rightarrow -p\sin t - q\cos t - 3p\cos t + 3q\sin t - 10p\sin t - 10q\cos t = 2\sin t - 3\cos t$	A1
$\Rightarrow -11p + 3q = 2$ and $3p + 11q = 3 \Rightarrow p = -0.1,\ q = 0.3$	dM1A1
GS: $x = Ae^{5t} + Be^{-2t} + 0.3\cos t - 0.1\sin t$	A1	CAO
Initial conditions $\Rightarrow A + B + 0.3 = 3.3 \Rightarrow A + B = 3$	B1
From $\dot{x} = 5Ae^{5t} - 2Be^{-2t} - 0.3\sin t - 0.1\cos t$: $5A - 2B - 0.1 = 0.9 \Rightarrow 5A - 2B = 1$	M1A1
$A = 1,\ B = 2 \Rightarrow x = e^{5t} + 2e^{-2t} + 0.3\cos t - 0.1\sin t$	A1 (11) Total 11	CAO

## Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(m-5)(m+2) = 0 \Rightarrow \text{CF}: x = Ae^{5t} + Be^{-2t}$ | M1A1 | |
| PI: $x = p\sin t + q\cos t \Rightarrow \dot{x} = p\cos t - q\sin t \Rightarrow \ddot{x} = -p\sin t - q\cos t$ | M1 | |
| $\Rightarrow -p\sin t - q\cos t - 3p\cos t + 3q\sin t - 10p\sin t - 10q\cos t = 2\sin t - 3\cos t$ | A1 | |
| $\Rightarrow -11p + 3q = 2$ and $3p + 11q = 3 \Rightarrow p = -0.1,\ q = 0.3$ | dM1A1 | |
| GS: $x = Ae^{5t} + Be^{-2t} + 0.3\cos t - 0.1\sin t$ | A1 | CAO |
| Initial conditions $\Rightarrow A + B + 0.3 = 3.3 \Rightarrow A + B = 3$ | B1 | |
| From $\dot{x} = 5Ae^{5t} - 2Be^{-2t} - 0.3\sin t - 0.1\cos t$: $5A - 2B - 0.1 = 0.9 \Rightarrow 5A - 2B = 1$ | M1A1 | |
| $A = 1,\ B = 2 \Rightarrow x = e^{5t} + 2e^{-2t} + 0.3\cos t - 0.1\sin t$ | A1 (11) Total 11 | CAO |

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9 Find the particular solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 10 x = 2 \sin t - 3 \cos t$$

given that, when $t = 0 , x = 3.3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.9$.

\hfill \mbox{\textit{CAIE FP1 2015 Q9 [11]}}

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